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+/*
+ * Copyright (c) 1992, 1993
+ *	The Regents of the University of California.  All rights reserved.
+ *
+ * This software was developed by the Computer Systems Engineering group
+ * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
+ * contributed to Berkeley.
+ *
+ * Redistribution and use in source and binary forms, with or without
+ * modification, are permitted provided that the following conditions
+ * are met:
+ * 1. Redistributions of source code must retain the above copyright
+ *    notice, this list of conditions and the following disclaimer.
+ * 2. Redistributions in binary form must reproduce the above copyright
+ *    notice, this list of conditions and the following disclaimer in the
+ *    documentation and/or other materials provided with the distribution.
+ * 3. All advertising materials mentioning features or use of this software
+ *    must display the following acknowledgement:
+ *	This product includes software developed by the University of
+ *	California, Berkeley and its contributors.
+ * 4. Neither the name of the University nor the names of its contributors
+ *    may be used to endorse or promote products derived from this software
+ *    without specific prior written permission.
+ *
+ * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
+ * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+ * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+ * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
+ * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
+ * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
+ * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
+ * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
+ * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
+ * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
+ * SUCH DAMAGE.
+ *
+ * from: Header: divrem.m4,v 1.4 92/06/25 13:23:57 torek Exp
+ * $NetBSD: divrem.m4,v 1.4 1997/10/09 10:07:54 lukem Exp $
+ */
+
+/*
+ * Division and remainder, from Appendix E of the Sparc Version 8
+ * Architecture Manual, with fixes from Gordon Irlam.
+ */
+
+#if defined(LIBC_SCCS) && !defined(lint)
+	.asciz "@(#)divrem.m4	8.1 (Berkeley) 6/4/93"
+#endif /* LIBC_SCCS and not lint */
+
+/*
+ * Input: dividend and divisor in %o0 and %o1 respectively.
+ *
+ * m4 parameters:
+ *  NAME	name of function to generate
+ *  OP		OP=div => %o0 / %o1; OP=rem => %o0 % %o1
+ *  S		S=true => signed; S=false => unsigned
+ *
+ * Algorithm parameters:
+ *  N		how many bits per iteration we try to get (4)
+ *  WORDSIZE	total number of bits (32)
+ *
+ * Derived constants:
+ *  TWOSUPN	2^N, for label generation (m4 exponentiation currently broken)
+ *  TOPBITS	number of bits in the top `decade' of a number
+ *
+ * Important variables:
+ *  Q		the partial quotient under development (initially 0)
+ *  R		the remainder so far, initially the dividend
+ *  ITER	number of main division loop iterations required;
+ *		equal to ceil(log2(quotient) / N).  Note that this
+ *		is the log base (2^N) of the quotient.
+ *  V		the current comparand, initially divisor*2^(ITER*N-1)
+ *
+ * Cost:
+ *  Current estimate for non-large dividend is
+ *	ceil(log2(quotient) / N) * (10 + 7N/2) + C
+ *  A large dividend is one greater than 2^(31-TOPBITS) and takes a
+ *  different path, as the upper bits of the quotient must be developed
+ *  one bit at a time.
+ */
+
+define(N, `4')
+define(TWOSUPN, `16')
+define(WORDSIZE, `32')
+define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))
+
+define(dividend, `%o0')
+define(divisor, `%o1')
+define(Q, `%o2')
+define(R, `%o3')
+define(ITER, `%o4')
+define(V, `%o5')
+
+/* m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d */
+define(T, `%g1')
+define(SC, `%g7')
+ifelse(S, `true', `define(SIGN, `%g6')')
+
+/*
+ * This is the recursive definition for developing quotient digits.
+ *
+ * Parameters:
+ *  $1	the current depth, 1 <= $1 <= N
+ *  $2	the current accumulation of quotient bits
+ *  N	max depth
+ *
+ * We add a new bit to $2 and either recurse or insert the bits in
+ * the quotient.  R, Q, and V are inputs and outputs as defined above;
+ * the condition codes are expected to reflect the input R, and are
+ * modified to reflect the output R.
+ */
+define(DEVELOP_QUOTIENT_BITS,
+`	! depth $1, accumulated bits $2
+	bl	L.$1.eval(TWOSUPN+$2)
+	srl	V,1,V
+	! remainder is positive
+	subcc	R,V,R
+	ifelse($1, N,
+	`	b	9f
+		add	Q, ($2*2+1), Q
+	', `	DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')')
+L.$1.eval(TWOSUPN+$2):
+	! remainder is negative
+	addcc	R,V,R
+	ifelse($1, N,
+	`	b	9f
+		add	Q, ($2*2-1), Q
+	', `	DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')')
+	ifelse($1, 1, `9:')')
+
+#include <machine/asm.h>
+#include <machine/trap.h>
+
+FUNC(NAME)
+ifelse(S, `true',
+`	! compute sign of result; if neither is negative, no problem
+	orcc	divisor, dividend, %g0	! either negative?
+	bge	2f			! no, go do the divide
+	ifelse(OP, `div',
+		`xor	divisor, dividend, SIGN',
+		`mov	dividend, SIGN')	! compute sign in any case
+	tst	divisor
+	bge	1f
+	tst	dividend
+	! divisor is definitely negative; dividend might also be negative
+	bge	2f			! if dividend not negative...
+	neg	divisor			! in any case, make divisor nonneg
+1:	! dividend is negative, divisor is nonnegative
+	neg	dividend		! make dividend nonnegative
+2:
+')
+	! Ready to divide.  Compute size of quotient; scale comparand.
+	orcc	divisor, %g0, V
+	bnz	1f
+	mov	dividend, R
+
+		! Divide by zero trap.  If it returns, return 0 (about as
+		! wrong as possible, but that is what SunOS does...).
+		t	ST_DIV0
+		retl
+		clr	%o0
+
+1:
+	cmp	R, V			! if divisor exceeds dividend, done
+	blu	Lgot_result		! (and algorithm fails otherwise)
+	clr	Q
+	sethi	%hi(1 << (WORDSIZE - TOPBITS - 1)), T
+	cmp	R, T
+	blu	Lnot_really_big
+	clr	ITER
+
+	! `Here the dividend is >= 2^(31-N) or so.  We must be careful here,
+	! as our usual N-at-a-shot divide step will cause overflow and havoc.
+	! The number of bits in the result here is N*ITER+SC, where SC <= N.
+	! Compute ITER in an unorthodox manner: know we need to shift V into
+	! the top decade: so do not even bother to compare to R.'
+	1:
+		cmp	V, T
+		bgeu	3f
+		mov	1, SC
+		sll	V, N, V
+		b	1b
+		inc	ITER
+
+	! Now compute SC.
+	2:	addcc	V, V, V
+		bcc	Lnot_too_big
+		inc	SC
+
+		! We get here if the divisor overflowed while shifting.
+		! This means that R has the high-order bit set.
+		! Restore V and subtract from R.
+		sll	T, TOPBITS, T	! high order bit
+		srl	V, 1, V		! rest of V
+		add	V, T, V
+		b	Ldo_single_div
+		dec	SC
+
+	Lnot_too_big:
+	3:	cmp	V, R
+		blu	2b
+		nop
+		be	Ldo_single_div
+		nop
+	/* NB: these are commented out in the V8-Sparc manual as well */
+	/* (I do not understand this) */
+	! V > R: went too far: back up 1 step
+	!	srl	V, 1, V
+	!	dec	SC
+	! do single-bit divide steps
+	!
+	! We have to be careful here.  We know that R >= V, so we can do the
+	! first divide step without thinking.  BUT, the others are conditional,
+	! and are only done if R >= 0.  Because both R and V may have the high-
+	! order bit set in the first step, just falling into the regular
+	! division loop will mess up the first time around.
+	! So we unroll slightly...
+	Ldo_single_div:
+		deccc	SC
+		bl	Lend_regular_divide
+		nop
+		sub	R, V, R
+		mov	1, Q
+		b	Lend_single_divloop
+		nop
+	Lsingle_divloop:
+		sll	Q, 1, Q
+		bl	1f
+		srl	V, 1, V
+		! R >= 0
+		sub	R, V, R
+		b	2f
+		inc	Q
+	1:	! R < 0
+		add	R, V, R
+		dec	Q
+	2:
+	Lend_single_divloop:
+		deccc	SC
+		bge	Lsingle_divloop
+		tst	R
+		b,a	Lend_regular_divide
+
+Lnot_really_big:
+1:
+	sll	V, N, V
+	cmp	V, R
+	bleu	1b
+	inccc	ITER
+	be	Lgot_result
+	dec	ITER
+
+	tst	R	! set up for initial iteration
+Ldivloop:
+	sll	Q, N, Q
+	DEVELOP_QUOTIENT_BITS(1, 0)
+Lend_regular_divide:
+	deccc	ITER
+	bge	Ldivloop
+	tst	R
+	bl,a	Lgot_result
+	! non-restoring fixup here (one instruction only!)
+ifelse(OP, `div',
+`	dec	Q
+', `	add	R, divisor, R
+')
+
+Lgot_result:
+ifelse(S, `true',
+`	! check to see if answer should be < 0
+	tst	SIGN
+	bl,a	1f
+	ifelse(OP, `div', `neg Q', `neg R')
+1:')
+	retl
+	ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')